What is the integral of (x-2)/((x^2+x+2)(x^2+2)) ?

The integral of (x-2)/((x^2+x+2)(x^2+2)) is

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integral of (x-2)/((x^2+x+2)(x^2+2))

\int \frac{x - 2}{( x ^{2} + x + 2 ) ( x ^{2} + 2 )} d x

- \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{3}{7} arctan (\frac{1}{7} (2 x + 1)) + \frac{1}{2} ln x^{2} + 2 + \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x - 2}{( x ^{2} + x + 2 ) ( x ^{2} + 2 )} d x

Take the partial fraction of

\frac{x - 2}{( x ^{2} + x + 2 ) ( x ^{2} + 2 )} : \frac{- x - 2}{x ^{2} + x + 2} + \frac{x + 1}{x ^{2} + 2}

= \int \frac{- x - 2}{x ^{2} + x + 2} + \frac{x + 1}{x ^{2} + 2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{- x - 2}{x ^{2} + x + 2} d x + \int \frac{x + 1}{x ^{2} + 2} d x

\int \frac{- x - 2}{x ^{2} + x + 2} d x = - \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{3}{7} arctan (\frac{1}{7} (2 x + 1))

\int \frac{x + 1}{x ^{2} + 2} d x = \frac{1}{2} ln x^{2} + 2 + \frac{1}{2} arctan (\frac{x}{2})

= - \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{3}{7} arctan (\frac{1}{7} (2 x + 1)) + \frac{1}{2} ln x^{2} + 2 + \frac{1}{2} arctan (\frac{x}{2})

Add a constant to the solution

= - \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{3}{7} arctan (\frac{1}{7} (2 x + 1)) + \frac{1}{2} ln x^{2} + 2 + \frac{1}{2} arctan (\frac{x}{2}) + C

\int_{0}^{4} π (2 x)^{2} d x

\frac{d}{d x} (x ln (x) - x + 1)

\int \frac{( 1 + 15 x )}{x} d x

\int \frac{x + 5}{x + 1} d x

x \to - 2 lim (x - 5)

What is the integral of (x-2)/((x^2+x+2)(x^2+2)) ?
The integral of (x-2)/((x^2+x+2)(x^2+2)) is